How do you prove that the function #f(x)=(x-5)/ |x-5|# is continuous everywhere but x=5?

1 Answer
Jim H
Nov 1, 2015

Use the definition of #abs u = { (-u,"if",u < 0),(u,"if",u >= 0) :}# to rewrite #f# as a piecewise function.

Explanation:

#f(x) = { (-1,"if",x<5),(1,"if",x>5) :}#

So, #f# is constant on #(-oo,5)#, hence continuous on #(-oo,5)#

and #f# is constant on #(5,oo)#, hence continuous on #(5,oo)#.

#f(5)# is not defined, so #f# is not continuous at #5#.

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